Consider the two curves `C_(1):y=1+cos x and C_(2): y=1 + cos (x-alpha)" for "alpha in (0,(pi)/(2))," where "x in [0,pi].` Also the area of the figure bounded by the curves `C_(1),C_(2), and x=0` is same as that of the figure bounded by `C_(2),y=1, and x=pi`.
The value of `alpha` is

Options

A. `(pi)/(4)`
B. `(pi)/(3)`
C. `(pi)/(6)`
D. `(pi)/(8)`

Explanation

To solve the problem, we need to find the value of \( \alpha \) such that the area bounded by the curves \( C_1: y = 1 + \cos x \) and \( C_2: y = 1 + \cos(x - \alpha) \) from \( x = 0 \) to the intersection point is equal to the area bounded by \( C_2 \), the line \( y = 1 \), and \( x = \pi \).

### Step-by-Step Solution:

1. **Identify the curves**:
- The first curve is \( C_1: y = 1 + \cos x \).
- The second curve is \( C_2: y = 1 + \cos(x - \alpha) \).

2. **Find the intersection points**:
- Set \( C_1 = C_2 \):
\[ 1 + \cos x = 1 + \cos(x - \alpha) \]
This simplifies to:
\[ \cos x = \cos(x - \alpha) \]
From this, we can derive:
\[ x = x - \alpha \quad \text{or} \quad x = 2\pi - (x - \alpha) \]
The first equation gives no new information, while the second leads to:
\[ 2x = 2\pi + \alpha \implies x = \pi + \frac{\alpha}{2} \]
However, since we are looking for the intersection in the interval \( [0, \pi] \), we will focus on the first equation.

3. **Calculate the area \( A_1 \)**:
- The area \( A_1 \) between the curves from \( x = 0 \) to \( x = \frac{\alpha}{2} \) is given by:
\[ A_1 = \int_0^{\frac{\alpha}{2}} (C_1 - C_2) \, dx = \int_0^{\frac{\alpha}{2}} \left( (1 + \cos x) - (1 + \cos(x - \alpha)) \right) \, dx \]
This simplifies to:
\[ A_1 = \int_0^{\frac{\alpha}{2}} \left( \cos x - \cos(x - \alpha) \right) \, dx \]

4. **Calculate the area \( A_2 \)**:
- The area \( A_2 \) between \( C_2 \) and the line \( y = 1 \) from \( x = \frac{\pi}{2} + \alpha \) to \( x = \pi \):
\[ A_2 = \int_{\frac{\pi}{2} + \alpha}^{\pi} (1 - C_2) \, dx = \int_{\frac{\pi}{2} + \alpha}^{\pi} \left( 1 - (1 + \cos(x - \alpha)) \right) \, dx \]
This simplifies to:
\[ A_2 = \int_{\frac{\pi}{2} + \alpha}^{\pi} -\cos(x - \alpha) \, dx \]

5. **Set the areas equal**:
- We have \( A_1 = A_2 \):
\[ \int_0^{\frac{\alpha}{2}} \left( \cos x - \cos(x - \alpha) \right) \, dx = \int_{\frac{\pi}{2} + \alpha}^{\pi} -\cos(x - \alpha) \, dx \]

6. **Evaluate the integrals**:
- The integral on the left can be evaluated using standard integration techniques.
- The integral on the right can be evaluated similarly.

7. **Solve for \( \alpha \)**:
- After evaluating both integrals and simplifying, we will find that:
\[ \sin\left(\frac{\alpha}{2}\right) = \frac{1}{2} \]
This gives:
\[ \frac{\alpha}{2} = \frac{\pi}{6} \implies \alpha = \frac{\pi}{3} \]

### Final Answer:
The value of \( \alpha \) is \( \frac{\pi}{3} \).